What does `layer.get_weights()` return?

Here is a working example.

import numpy as np
from  keras.models import Sequential
from keras.layers import Dense, Dropout, Activation, Flatten

X_train=np.random.rand(1,10)
Y_train=2*X_train
input_dim = X_train.shape[1]
model=Sequential()
model.add(Dense(20, input_dim=10))
model.add(Dense(10, activation='softmax'))
weight_origin_0=model.layers[0].get_weights()[0]
weight_origin_1=model.layers[1].get_weights()[0]
model.compile(loss='categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
model.fit(X_train, Y_train, batch_size=1, nb_epoch=10, verbose=1)
print(weight_origin_0-model.layers[0].get_weights()[0])  #the first layer
print(weight_origin_1-model.layers[1].get_weights()[0])  #the second layer

For the question of layer.get_weights():

I ran some tests on this issue and checked the source codes. I found that the Dense layer is a subclass of Layer and its weights, which is a type of python list has two elements weight of the layer stored at layer.get_weights()[0] and the bias is stored at layer.get_weights()[1].

There's one thing to note that, bias can be disabled during defining the layer: model.add(Dense(503,init='normal',activation='relu',bias=False)). In that case, the list layer.get_weights() has only one element. If you set the bias attribute as False after defining it, there will still be an element for bias and it would be updated after you fitting the model.

For the question of not updating:

I set up a Sequential model with only one dense layer:

def mlp_2(): model=Sequential() model.add(Dense(10, input_dim=784, activation='softmax', bias =False)) return model

Then I use the same way above to compile and fit it. This is what I got:

It still seems not update the weight, however, we can tell the weight is definately changed. Because the accuracy is increasing. I think the only explanation is updates on the first dense layer (which you define input_dim) is too small for Keras to printout. I didn't check the more precise value of the weights, it would be great if someone could confrim it.